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Measure Theory and Lebesgue Integration

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Measure Theory and Lebesgue Integration T. Muthukumar [email protected] April 12, 2018 ii Contents Notationsv 1 Introduction1 1.1 Riemann Integration and its Inadequacy............1 1.1.1 Limit and Integral: Interchange.............3 1.1.2 Differentiation and Integration: Duality.........5 1.2 Motivating Lebesgue Integral and Measure...........6 2 Lebesgue Measure on Rn 9 2.1 Introduction............................9 2.2 Outer measure.......................... 12 2.2.1 Abstract Set-up...................... 22 2.3 Measurable Sets.......................... 22 2.3.1 Abstract Set-up...................... 32 2.4 Measurable Functions....................... 34 2.5 Littlewood's Three Principles.................. 43 2.5.1 First Principle....................... 44 2.5.2 Third Principle...................... 45 2.5.3 Second Principle..................... 49 2.6 Jordan Content or Measure................... 50 3 Lebesgue Integration 55 3.1 Simple Functions......................... 55 3.2 Bounded Function With Finite Measure Support........ 58 3.3 Non-negative Functions...................... 62 3.4 General Integrable Functions................... 67 3.5 Order of Integration....................... 75 3.6 Lp Spaces............................. 78 iii CONTENTS iv 3.7 Invariance of Lebesgue Integral................. 87 4 Duality of Differentiation and Integration 89 4.1 Monotone Functions....................... 90 4.2 Bounded Variation Functions.................. 97 4.3 Derivative of an Integral..................... 106 4.4 Absolute Continuity and FTC.................. 111 Appendices 119 A Cantor Set and Cantor Function 121 Bibliography 127 Notations Symbols 2S will denote the power set, the set of all subsets, of a set S L(Rn) the class of all Lebesgue measurable subsets of Rn C denotes the plane of complex numbers Q denotes the set of all rationals in R Qn set of all vectors in Rn with each coordinate being rational number R denotes the real line Rn denotes the Euclidean space of dimension n Function Spaces R([a; b]) denotes the space of all Riemann integrable functions on the interval [a; b] Lip(E) denotes the space of all Lipschitz functions on E AC(E) denotes the space of all absolutely continuous functions on E BV (E) denotes the space of all bounded variation functions on E C(X) the class of all real-valued continuous functions on X C0(X) denotes the space of all continuous functions vanishing at 1 on X Cc(X) denotes the space of all compactly supported continuous functions on X v NOTATIONS vi Lp(E) denotes the space of all measurable p-integrable functions on E M(Rn) the class of all finite a.e. real valued Lebesgue measurable functions on Rn General Conventions Br(x) will denote the closed ball of radius r and centre at x Ec will denote the set complement of E ⊂ S, S n E Chapter 1 Introduction 1.1 Riemann Integration and its Inadequacy Let f :[a; b] ! R be a bounded function. Let P be the partition of the interval [a; b], a = x0 ≤ x1 ≤ ::: ≤ xk = b. For i = 0; 1; 2; : : : ; k, let Mi(P ) = sup f(x) and mi(P ) = inf f(x): x2[x ;x ] x2[xi−1;xi] i−1 i The upper Riemann sum of f with respect to the given partition P is, k X U(P; f) = Mi(P )(xi − xi−1) i=1 and the lower Riemann sum of f with respect to the given partition P is, k X L(P; f) = mi(P )(xi − xi−1): i=1 We say the bounded function f is Riemann integrable on [a; b] if the infimum of upper sum and supremum of lower sum, over all partitions P of [a; b], coincide and is denoted as Z b f(x) dx := inf U(P; f) = sup L(P; f): a P P If f = u + iv is a bounded complex-valued function on [a; b], then f is said to be Riemann integrable if its real and imaginary parts are Riemann 1 CHAPTER 1. INTRODUCTION 2 integral and Z b Z b Z b f(x) dx = u(x) dx + i v(x) dx: a a a If either f is unbounded or the domain [a; b] is not finite then its corre- sponding integral, called as improper integral, is defined in terms of limits of Riemann integrable functions, whenever possible. Exercise 1. Every Riemann integrable function1 is bounded. Let R([a; b]) denote the space of all Riemann integrable functions on [a; b]. The space R([a; b]) forms a vector space over R (or C). It is closed under composition, if it makes sense. Theorem 1.1.1. If f is continuous on [a; b], then f 2 R([a; b]). In fact even piecewise continuity is sufficient for Riemann integrability. Theorem 1.1.2. If f is continuous except at finitely many points of [a; b] (piecewise continu- ous), then f 2 R([a; b]). But there are functions which has discontinuity at countably many points and are still in R([a; b]). Example 1.1. Consider the function 8 1 1 >1 if k+1 < x ≤ k and k is odd < 1 1 f(x) = 0 if k+1 < x ≤ k and k is even :>0 x = 0 which has discontinuities at x = 0 and x = 1=k, for k = 1; 2;:::. It can be shown that f 2 R([0; 1]). Theorem 1.1.3. If f is bounded monotonic on [a; b] then f 2 R([a; b]). In fact, one can construct functions whose set of discontinuities are `dense' in [0; 1]. 1here by Riemann integrable we mean the upper sum and lower sum coincide and are finite CHAPTER 1. INTRODUCTION 3 1 Example 1.2. For instance, let frkg1 denote a countable dense subset of [0; 1] (for instance, Q) and define 1 X 1 f(x) = H(x − r ) k2 k k=1 where H : R ! R is defined as ( 1 if x ≥ 0 H(x) = 0 if x < 0: The function f is discontinuous at all the points rk and can be shown to be in R([0; 1]), because it is bounded and monotone. Theorem 1.1.4. If f 2 R([a; b]) then f is continuous on a dense subset of [a; b]. Example 1.3. An example of a function f : [0; 1] ! R which is not Riemann integrable is ( 1 x 2 f(x) = Q 0 x 2 [0; 1] n Q: A necessary and sufficient condition of Riemann integrability is given by Theorem 3.0.1. Thus, even to characterise the class of Riemann integrable functions, we need to have the notion of length (\measure") (at least measure zero). 1.1.1 Limit and Integral: Interchange Let us consider a sequence of functions ffkg ⊂ R([a; b]) and define f(x) := limk!1 fk(x), assuming that the limit exists for every x 2 [a; b]. Does f 2 R([a; b])? The answer is a \no", as seen in example below. Example 1.4. Fix an enumeration (order) of the set of rationals in [0; 1]. Let the finite set rk denote the first k elements of the set of rationals in [0; 1]. Define the sequence of functions ( 1 if x 2 rk fk(x) = 0 otherwise: CHAPTER 1. INTRODUCTION 4 Each fk 2 R([0; 1]), since it has discontinuity at k (finite) number of points. The point-wise limit of fk, f = limk!1 fk, is ( 1 x 2 f(x) = Q 0 x 2 [0; 1] n Q which we have seen above is not Riemann integrable. Thus, the space R([a; b]) is not \complete" under point-wise limit. How- ever, R([a; b]) is complete under uniform convergence. A related question is if the limit f 2 R([a; b]), is the Riemann integral of f the limit of the Riemann integrals of fk, i.e., can we say Z b Z b f(x) dx = lim fk(x) dx? a k!1 a The answer is a \no" again. Example 1.5. Consider the functions ( k x 2 (0; 1=k) fk(x) = 0 otherwise Then f(x) = limk fk(x) = 0. Note that Z fk(x) dx = 1 8k; R but R f(x) dx = 0. R The interchange becomes possible under uniform convergence. Theorem 1.1.5. Let ffkg ⊂ R([a; b]) and fk(x) ! f(x) uniformly in [a; b]. Then f 2 R([a; b]) and Z b Z b f = lim fk: a k!1 a But uniform convergence is too demanding in practice. The following more general result for interchanging limit and integral will be proved in this write-up. CHAPTER 1. INTRODUCTION 5 Theorem 1.1.6. Let ffkg ⊂ R([a; b]) and f 2 R([a; b]). Also, let fk(x) ! f(x) point-wise and fk are uniformly bounded. Then Z b Z b lim fk = f: k!1 a a The proof of above theorem is not elementary, thus in classical analysis we always prove the result for uniform convergence. Observe the hypothesis of integrability on f in the above theorem. 1.1.2 Differentiation and Integration: Duality An observation we make, once we have Riemann integration, is about the dual nature of differentiation and integration. Thus, one asks the following two questions: 1. (Derivative of an integral) For which class of functions can we say d Z x f(t) dt = f(x)? dx a 2. (Integral of a derivative) For which class of functions can we say Z b f 0(x) dx = f(b) − f(a)? a To answer the first question, for any f 2 R([a; b]), let us define the function Z x F (x) := f(t) dt: a Exercise 2.
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